Question

Question #44d73

Answer 1

Here's what I got.

Explanation:

Start by calculating the energy of a photon of wavelength `"300 nm"` by using the Planck - Einstein relation

`color(blue)(ul(color(black)(E = h * c/(lamda))))`

Here

• `E` is the energy of the photon
• `h` is Planck's constant, equal to `6.626 * 10^(-34)"J s"`
• `c` is the speed of light in a vacuum, usually given as `3 * 10^8` `"m s"^(-1)`
• `lamda` is the wavelength of the photon

Plug in your values to find--do not forget to convert the wavelength to meters

`E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * (3 * 10^8color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(300 * 10^(-9)color(red)(cancel(color(black)("m"))))`

`E = 6.63 * 10^(-19)` `"J"`

At this point, you can convert the work function of the metal from electronvolts to joules or the energy of the incoming photon from joules to electronvolts by using the fact that

`"1 eV" = 1.9 * 10^(-19)` `"J"`

To make the subsequent calculations easier, I'll go with the latter.

`6.63 * color(blue)(cancel(color(black)(10^(-19)))) color(red)(cancel(color(black)("J"))) * "1 eV"/(1.6 * color(blue)(cancel(color(black)(10^(-19))))color(red)(cancel(color(black)("J")))) = "4.14 eV"`

Now, the idea here is that the work function of the metal, `Psi` tells you the minimum energy needed in order to remove an electron from the surface of the metal to a point located immediately outside the surface.

In this case, you know that you need `"2.46 eV"` in order to remove an electron from the surface of sodium metal.

The difference between the energy of the incoming photon and the work function of the metal will give you the kinetic energy of the electron `->` think the photoelectric effect here.

`K_"E" = E - phi`

In your case, you have

`K_"E" = "4.14 eV" - "2.46 eV" = color(darkgreen)(ul(color(black)("1.7 eV")))`

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the wavelength of the incoming photon.